June 2016 Paper 2 Q21
21 Solve \(\quad \dfrac{3}{x - 2} + \dfrac{2}{x - 1} = 5\)
Do not use trial and improvement.
Write your solutions to 3 significant figures. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3(x - 1)\) or \(3x - 3\) or \(2(x - 2)\) or \(2x - 4\) | M1 | |
| \(3x - 3\) and \(2x - 4\) or \(5x - 7\) | M1 | Implies M1 M1 |
| \(5(x - 1)(x - 2)\) or \(5(x^2 - 2x - x + 2)\) or \(5x^2 - 15x + 10\) or \((x - 1)(x - 2)\) expanded and multiplied by 5 | M1 | oe Allow one error in four term expansion of \(5(x - 1)(x - 2)\) Implied by \(5(x^2 - 3x + k)\) or \(5(ax^2 - 3x + 2)\) |
| \(5x^2 - 20x + 17\ (= 0)\) | M1dep | dep on 3rd M1 oe 3-term quadratic equation eg \(5x^2 - 20x = -17\) Correctly collects terms in their expansion |
| \(\dfrac{--20 \pm \sqrt{(-20)^2 - 4 \times 5 \times 17}}{2 \times 5}\) or \(\dfrac{10 \pm \sqrt{15}}{5}\) or \((x - 2)^2 - 4 = -\dfrac{17}{5}\) or \(5[(x - 2)^2 - 4] = -17\) | M1 | oe Correct use of quadratic formula for their 3-term quadratic eg \((-20)^2\) can be \(20^2\) or correct factorisation of their 3-term quadratic or attempt to complete the square for their 3-term quadratic Must be correct up to form \((x - a)^2 + b = c\) or \(k[(x - d)^2 + e] = f\) |
| 1.23 and 2.77 | A1 | Must be 3 significant figures |
Additional guidance
| For A1, the word ‘and’ is not needed eg 1.23, 2.77 (with method seen) | M5 A1 |
| Brackets may be recovered throughout | |
| 5th M1 may be implied by solutions of their quadratic equation seen | |
| M0 M0 M0 M0 M1 A0 is possible if they have a 3-term quadratic equation | |
| Answers only | Zero |